\(\int \cos ^3(c+d x) (a+b \sin (c+d x))^2 (A+B \sin (c+d x)) \, dx\) [1534]

Optimal result

Integrand size = 31, antiderivative size = 132 \[ \int \cos ^3(c+d x) (a+b \sin (c+d x))^2 (A+B \sin (c+d x)) \, dx=-\frac {\left (a^2-b^2\right ) (A b-a B) (a+b \sin (c+d x))^3}{3 b^4 d}+\frac {\left (2 a A b-3 a^2 B+b^2 B\right ) (a+b \sin (c+d x))^4}{4 b^4 d}-\frac {(A b-3 a B) (a+b \sin (c+d x))^5}{5 b^4 d}-\frac {B (a+b \sin (c+d x))^6}{6 b^4 d} \] Output:

-1/3*(a^2-b^2)*(A*b-B*a)*(a+b*sin(d*x+c))^3/b^4/d+1/4*(2*A*a*b-3*B*a^2+B*b 
^2)*(a+b*sin(d*x+c))^4/b^4/d-1/5*(A*b-3*B*a)*(a+b*sin(d*x+c))^5/b^4/d-1/6* 
B*(a+b*sin(d*x+c))^6/b^4/d
 

Mathematica [A] (verified)

Time = 0.26 (sec) , antiderivative size = 111, normalized size of antiderivative = 0.84 \[ \int \cos ^3(c+d x) (a+b \sin (c+d x))^2 (A+B \sin (c+d x)) \, dx=\frac {(a+b \sin (c+d x))^3 \left (-2 a^2 A b+20 A b^3+a^3 B-5 a b^2 B+3 b \left (2 a A b-a^2 B+5 b^2 B\right ) \sin (c+d x)-6 b^2 (2 A b-a B) \sin ^2(c+d x)-10 b^3 B \sin ^3(c+d x)\right )}{60 b^4 d} \] Input:

Integrate[Cos[c + d*x]^3*(a + b*Sin[c + d*x])^2*(A + B*Sin[c + d*x]),x]
 

Output:

((a + b*Sin[c + d*x])^3*(-2*a^2*A*b + 20*A*b^3 + a^3*B - 5*a*b^2*B + 3*b*( 
2*a*A*b - a^2*B + 5*b^2*B)*Sin[c + d*x] - 6*b^2*(2*A*b - a*B)*Sin[c + d*x] 
^2 - 10*b^3*B*Sin[c + d*x]^3))/(60*b^4*d)
 

Rubi [A] (verified)

Time = 0.37 (sec) , antiderivative size = 115, normalized size of antiderivative = 0.87, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.161, Rules used = {3042, 3316, 27, 652, 2009}

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

\(\displaystyle \int \cos ^3(c+d x) (a+b \sin (c+d x))^2 (A+B \sin (c+d x)) \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \cos (c+d x)^3 (a+b \sin (c+d x))^2 (A+B \sin (c+d x))dx\)

\(\Big \downarrow \) 3316

\(\displaystyle \frac {\int \frac {(a+b \sin (c+d x))^2 (A b+B \sin (c+d x) b) \left (b^2-b^2 \sin ^2(c+d x)\right )}{b}d(b \sin (c+d x))}{b^3 d}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\int (a+b \sin (c+d x))^2 (A b+B \sin (c+d x) b) \left (b^2-b^2 \sin ^2(c+d x)\right )d(b \sin (c+d x))}{b^4 d}\)

\(\Big \downarrow \) 652

\(\displaystyle \frac {\int \left (-B (a+b \sin (c+d x))^5+(3 a B-A b) (a+b \sin (c+d x))^4+\left (-3 B a^2+2 A b a+b^2 B\right ) (a+b \sin (c+d x))^3+\left (a^2-b^2\right ) (a B-A b) (a+b \sin (c+d x))^2\right )d(b \sin (c+d x))}{b^4 d}\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {\frac {1}{4} \left (-3 a^2 B+2 a A b+b^2 B\right ) (a+b \sin (c+d x))^4-\frac {1}{3} \left (a^2-b^2\right ) (A b-a B) (a+b \sin (c+d x))^3-\frac {1}{5} (A b-3 a B) (a+b \sin (c+d x))^5-\frac {1}{6} B (a+b \sin (c+d x))^6}{b^4 d}\)

Input:

Int[Cos[c + d*x]^3*(a + b*Sin[c + d*x])^2*(A + B*Sin[c + d*x]),x]
 

Output:

(-1/3*((a^2 - b^2)*(A*b - a*B)*(a + b*Sin[c + d*x])^3) + ((2*a*A*b - 3*a^2 
*B + b^2*B)*(a + b*Sin[c + d*x])^4)/4 - ((A*b - 3*a*B)*(a + b*Sin[c + d*x] 
)^5)/5 - (B*(a + b*Sin[c + d*x])^6)/6)/(b^4*d)
 

Defintions of rubi rules used

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 652

Int[((d_.) + (e_.)*(x_))^(m_.)*((f_.) + (g_.)*(x_))^(n_.)*((a_) + (c_.)*(x_ 
)^2)^(p_.), x_Symbol] :> Int[ExpandIntegrand[(d + e*x)^m*(f + g*x)^n*(a + c 
*x^2)^p, x], x] /; FreeQ[{a, c, d, e, f, g, m, n}, x] && IGtQ[p, 0]
 

rule 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

rule 3042

Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 3316

Int[cos[(e_.) + (f_.)*(x_)]^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_ 
.)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Simp[1/(b^p* 
f)   Subst[Int[(a + x)^m*(c + (d/b)*x)^n*(b^2 - x^2)^((p - 1)/2), x], x, b* 
Sin[e + f*x]], x] /; FreeQ[{a, b, c, d, e, f, m, n}, x] && IntegerQ[(p - 1) 
/2] && NeQ[a^2 - b^2, 0]
 

Maple [A] (verified)

Time = 78.70 (sec) , antiderivative size = 130, normalized size of antiderivative = 0.98

method	result	size
derivativedivides	\(-\frac {\frac {B \,b^{2} \sin \left (d x +c \right )^{6}}{6}+\frac {\left (b^{2} A +2 a b B \right ) \sin \left (d x +c \right )^{5}}{5}+\frac {\left (\left (a^{2}-b^{2}\right ) B +2 A a b \right ) \sin \left (d x +c \right )^{4}}{4}+\frac {\left (-2 a b B +\left (a^{2}-b^{2}\right ) A \right ) \sin \left (d x +c \right )^{3}}{3}+\frac {\left (-2 A a b -a^{2} B \right ) \sin \left (d x +c \right )^{2}}{2}-\sin \left (d x +c \right ) A \,a^{2}}{d}\)	\(130\)
default	\(-\frac {\frac {B \,b^{2} \sin \left (d x +c \right )^{6}}{6}+\frac {\left (b^{2} A +2 a b B \right ) \sin \left (d x +c \right )^{5}}{5}+\frac {\left (\left (a^{2}-b^{2}\right ) B +2 A a b \right ) \sin \left (d x +c \right )^{4}}{4}+\frac {\left (-2 a b B +\left (a^{2}-b^{2}\right ) A \right ) \sin \left (d x +c \right )^{3}}{3}+\frac {\left (-2 A a b -a^{2} B \right ) \sin \left (d x +c \right )^{2}}{2}-\sin \left (d x +c \right ) A \,a^{2}}{d}\)	\(130\)
parallelrisch	\(\frac {\left (-240 A a b -120 a^{2} B -45 B \,b^{2}\right ) \cos \left (2 d x +2 c \right )+\left (80 A \,a^{2}-20 b^{2} A -40 a b B \right ) \sin \left (3 d x +3 c \right )+\left (-60 A a b -30 a^{2} B \right ) \cos \left (4 d x +4 c \right )+\left (-12 b^{2} A -24 a b B \right ) \sin \left (5 d x +5 c \right )+5 B \cos \left (6 d x +6 c \right ) b^{2}+\left (720 A \,a^{2}+120 b^{2} A +240 a b B \right ) \sin \left (d x +c \right )+300 A a b +150 a^{2} B +40 B \,b^{2}}{960 d}\)	\(164\)
risch	\(\frac {3 \sin \left (d x +c \right ) A \,a^{2}}{4 d}+\frac {\sin \left (d x +c \right ) b^{2} A}{8 d}+\frac {\sin \left (d x +c \right ) a b B}{4 d}+\frac {B \,b^{2} \cos \left (6 d x +6 c \right )}{192 d}-\frac {\sin \left (5 d x +5 c \right ) b^{2} A}{80 d}-\frac {\sin \left (5 d x +5 c \right ) a b B}{40 d}-\frac {a \cos \left (4 d x +4 c \right ) A b}{16 d}-\frac {a^{2} \cos \left (4 d x +4 c \right ) B}{32 d}+\frac {\sin \left (3 d x +3 c \right ) A \,a^{2}}{12 d}-\frac {\sin \left (3 d x +3 c \right ) b^{2} A}{48 d}-\frac {\sin \left (3 d x +3 c \right ) a b B}{24 d}-\frac {\cos \left (2 d x +2 c \right ) A a b}{4 d}-\frac {\cos \left (2 d x +2 c \right ) a^{2} B}{8 d}-\frac {3 \cos \left (2 d x +2 c \right ) B \,b^{2}}{64 d}\)	\(240\)
norman	\(\frac {\frac {\left (8 A a b +4 a^{2} B +4 B \,b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}}{d}+\frac {\left (8 A a b +4 a^{2} B +4 B \,b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{8}}{d}+\frac {2 \left (11 A \,a^{2}+4 b^{2} A +8 a b B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{3 d}+\frac {2 \left (11 A \,a^{2}+4 b^{2} A +8 a b B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{9}}{3 d}+\frac {4 \left (15 A \,a^{2}+2 b^{2} A +4 a b B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{5 d}+\frac {4 \left (15 A \,a^{2}+2 b^{2} A +4 a b B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{5 d}+\frac {2 \left (2 A a b +a^{2} B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}{d}+\frac {2 \left (2 A a b +a^{2} B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{10}}{d}+\frac {4 \left (6 A a b +3 a^{2} B -2 B \,b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}}{3 d}+\frac {2 A \,a^{2} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d}+\frac {2 A \,a^{2} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{11}}{d}}{\left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{6}}\)	\(346\)
orering	\(\text {Expression too large to display}\)	\(2462\)

Input:

int(cos(d*x+c)^3*(a+b*sin(d*x+c))^2*(A+B*sin(d*x+c)),x,method=_RETURNVERBO 
SE)
 

Output:

-1/d*(1/6*B*b^2*sin(d*x+c)^6+1/5*(A*b^2+2*B*a*b)*sin(d*x+c)^5+1/4*((a^2-b^ 
2)*B+2*A*a*b)*sin(d*x+c)^4+1/3*(-2*a*b*B+(a^2-b^2)*A)*sin(d*x+c)^3+1/2*(-2 
*A*a*b-B*a^2)*sin(d*x+c)^2-sin(d*x+c)*A*a^2)
 

Fricas [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 120, normalized size of antiderivative = 0.91 \[ \int \cos ^3(c+d x) (a+b \sin (c+d x))^2 (A+B \sin (c+d x)) \, dx=\frac {10 \, B b^{2} \cos \left (d x + c\right )^{6} - 15 \, {\left (B a^{2} + 2 \, A a b + B b^{2}\right )} \cos \left (d x + c\right )^{4} - 4 \, {\left (3 \, {\left (2 \, B a b + A b^{2}\right )} \cos \left (d x + c\right )^{4} - 10 \, A a^{2} - 4 \, B a b - 2 \, A b^{2} - {\left (5 \, A a^{2} + 2 \, B a b + A b^{2}\right )} \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )}{60 \, d} \] Input:

integrate(cos(d*x+c)^3*(a+b*sin(d*x+c))^2*(A+B*sin(d*x+c)),x, algorithm="f 
ricas")
 

Output:

1/60*(10*B*b^2*cos(d*x + c)^6 - 15*(B*a^2 + 2*A*a*b + B*b^2)*cos(d*x + c)^ 
4 - 4*(3*(2*B*a*b + A*b^2)*cos(d*x + c)^4 - 10*A*a^2 - 4*B*a*b - 2*A*b^2 - 
 (5*A*a^2 + 2*B*a*b + A*b^2)*cos(d*x + c)^2)*sin(d*x + c))/d
 

Sympy [A] (verification not implemented)

Time = 0.36 (sec) , antiderivative size = 228, normalized size of antiderivative = 1.73 \[ \int \cos ^3(c+d x) (a+b \sin (c+d x))^2 (A+B \sin (c+d x)) \, dx=\begin {cases} \frac {2 A a^{2} \sin ^{3}{\left (c + d x \right )}}{3 d} + \frac {A a^{2} \sin {\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{d} - \frac {A a b \cos ^{4}{\left (c + d x \right )}}{2 d} + \frac {2 A b^{2} \sin ^{5}{\left (c + d x \right )}}{15 d} + \frac {A b^{2} \sin ^{3}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{3 d} - \frac {B a^{2} \cos ^{4}{\left (c + d x \right )}}{4 d} + \frac {4 B a b \sin ^{5}{\left (c + d x \right )}}{15 d} + \frac {2 B a b \sin ^{3}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{3 d} + \frac {B b^{2} \sin ^{6}{\left (c + d x \right )}}{12 d} + \frac {B b^{2} \sin ^{4}{\left (c + d x \right )} \cos ^{2}{\left (c + d x \right )}}{4 d} & \text {for}\: d \neq 0 \\x \left (A + B \sin {\left (c \right )}\right ) \left (a + b \sin {\left (c \right )}\right )^{2} \cos ^{3}{\left (c \right )} & \text {otherwise} \end {cases} \] Input:

integrate(cos(d*x+c)**3*(a+b*sin(d*x+c))**2*(A+B*sin(d*x+c)),x)
 

Output:

Piecewise((2*A*a**2*sin(c + d*x)**3/(3*d) + A*a**2*sin(c + d*x)*cos(c + d* 
x)**2/d - A*a*b*cos(c + d*x)**4/(2*d) + 2*A*b**2*sin(c + d*x)**5/(15*d) + 
A*b**2*sin(c + d*x)**3*cos(c + d*x)**2/(3*d) - B*a**2*cos(c + d*x)**4/(4*d 
) + 4*B*a*b*sin(c + d*x)**5/(15*d) + 2*B*a*b*sin(c + d*x)**3*cos(c + d*x)* 
*2/(3*d) + B*b**2*sin(c + d*x)**6/(12*d) + B*b**2*sin(c + d*x)**4*cos(c + 
d*x)**2/(4*d), Ne(d, 0)), (x*(A + B*sin(c))*(a + b*sin(c))**2*cos(c)**3, T 
rue))
 

Maxima [A] (verification not implemented)

Time = 0.03 (sec) , antiderivative size = 128, normalized size of antiderivative = 0.97 \[ \int \cos ^3(c+d x) (a+b \sin (c+d x))^2 (A+B \sin (c+d x)) \, dx=-\frac {10 \, B b^{2} \sin \left (d x + c\right )^{6} + 12 \, {\left (2 \, B a b + A b^{2}\right )} \sin \left (d x + c\right )^{5} + 15 \, {\left (B a^{2} + 2 \, A a b - B b^{2}\right )} \sin \left (d x + c\right )^{4} - 60 \, A a^{2} \sin \left (d x + c\right ) + 20 \, {\left (A a^{2} - 2 \, B a b - A b^{2}\right )} \sin \left (d x + c\right )^{3} - 30 \, {\left (B a^{2} + 2 \, A a b\right )} \sin \left (d x + c\right )^{2}}{60 \, d} \] Input:

integrate(cos(d*x+c)^3*(a+b*sin(d*x+c))^2*(A+B*sin(d*x+c)),x, algorithm="m 
axima")
 

Output:

-1/60*(10*B*b^2*sin(d*x + c)^6 + 12*(2*B*a*b + A*b^2)*sin(d*x + c)^5 + 15* 
(B*a^2 + 2*A*a*b - B*b^2)*sin(d*x + c)^4 - 60*A*a^2*sin(d*x + c) + 20*(A*a 
^2 - 2*B*a*b - A*b^2)*sin(d*x + c)^3 - 30*(B*a^2 + 2*A*a*b)*sin(d*x + c)^2 
)/d
                                                                                    
                                                                                    
 

Giac [A] (verification not implemented)

Time = 0.37 (sec) , antiderivative size = 168, normalized size of antiderivative = 1.27 \[ \int \cos ^3(c+d x) (a+b \sin (c+d x))^2 (A+B \sin (c+d x)) \, dx=-\frac {10 \, B b^{2} \sin \left (d x + c\right )^{6} + 24 \, B a b \sin \left (d x + c\right )^{5} + 12 \, A b^{2} \sin \left (d x + c\right )^{5} + 15 \, B a^{2} \sin \left (d x + c\right )^{4} + 30 \, A a b \sin \left (d x + c\right )^{4} - 15 \, B b^{2} \sin \left (d x + c\right )^{4} + 20 \, A a^{2} \sin \left (d x + c\right )^{3} - 40 \, B a b \sin \left (d x + c\right )^{3} - 20 \, A b^{2} \sin \left (d x + c\right )^{3} - 30 \, B a^{2} \sin \left (d x + c\right )^{2} - 60 \, A a b \sin \left (d x + c\right )^{2} - 60 \, A a^{2} \sin \left (d x + c\right )}{60 \, d} \] Input:

integrate(cos(d*x+c)^3*(a+b*sin(d*x+c))^2*(A+B*sin(d*x+c)),x, algorithm="g 
iac")
 

Output:

-1/60*(10*B*b^2*sin(d*x + c)^6 + 24*B*a*b*sin(d*x + c)^5 + 12*A*b^2*sin(d* 
x + c)^5 + 15*B*a^2*sin(d*x + c)^4 + 30*A*a*b*sin(d*x + c)^4 - 15*B*b^2*si 
n(d*x + c)^4 + 20*A*a^2*sin(d*x + c)^3 - 40*B*a*b*sin(d*x + c)^3 - 20*A*b^ 
2*sin(d*x + c)^3 - 30*B*a^2*sin(d*x + c)^2 - 60*A*a*b*sin(d*x + c)^2 - 60* 
A*a^2*sin(d*x + c))/d
 

Mupad [B] (verification not implemented)

Time = 34.00 (sec) , antiderivative size = 127, normalized size of antiderivative = 0.96 \[ \int \cos ^3(c+d x) (a+b \sin (c+d x))^2 (A+B \sin (c+d x)) \, dx=\frac {{\sin \left (c+d\,x\right )}^2\,\left (\frac {B\,a^2}{2}+A\,b\,a\right )-{\sin \left (c+d\,x\right )}^5\,\left (\frac {A\,b^2}{5}+\frac {2\,B\,a\,b}{5}\right )+{\sin \left (c+d\,x\right )}^3\,\left (-\frac {A\,a^2}{3}+\frac {2\,B\,a\,b}{3}+\frac {A\,b^2}{3}\right )-{\sin \left (c+d\,x\right )}^4\,\left (\frac {B\,a^2}{4}+\frac {A\,a\,b}{2}-\frac {B\,b^2}{4}\right )-\frac {B\,b^2\,{\sin \left (c+d\,x\right )}^6}{6}+A\,a^2\,\sin \left (c+d\,x\right )}{d} \] Input:

int(cos(c + d*x)^3*(A + B*sin(c + d*x))*(a + b*sin(c + d*x))^2,x)
 

Output:

(sin(c + d*x)^2*((B*a^2)/2 + A*a*b) - sin(c + d*x)^5*((A*b^2)/5 + (2*B*a*b 
)/5) + sin(c + d*x)^3*((A*b^2)/3 - (A*a^2)/3 + (2*B*a*b)/3) - sin(c + d*x) 
^4*((B*a^2)/4 - (B*b^2)/4 + (A*a*b)/2) - (B*b^2*sin(c + d*x)^6)/6 + A*a^2* 
sin(c + d*x))/d
 

Reduce [B] (verification not implemented)

Time = 0.16 (sec) , antiderivative size = 110, normalized size of antiderivative = 0.83 \[ \int \cos ^3(c+d x) (a+b \sin (c+d x))^2 (A+B \sin (c+d x)) \, dx=\frac {\sin \left (d x +c \right ) \left (-10 \sin \left (d x +c \right )^{5} b^{3}-36 \sin \left (d x +c \right )^{4} a \,b^{2}-45 \sin \left (d x +c \right )^{3} a^{2} b +15 \sin \left (d x +c \right )^{3} b^{3}-20 \sin \left (d x +c \right )^{2} a^{3}+60 \sin \left (d x +c \right )^{2} a \,b^{2}+90 \sin \left (d x +c \right ) a^{2} b +60 a^{3}\right )}{60 d} \] Input:

int(cos(d*x+c)^3*(a+b*sin(d*x+c))^2*(A+B*sin(d*x+c)),x)
 

Output:

(sin(c + d*x)*( - 10*sin(c + d*x)**5*b**3 - 36*sin(c + d*x)**4*a*b**2 - 45 
*sin(c + d*x)**3*a**2*b + 15*sin(c + d*x)**3*b**3 - 20*sin(c + d*x)**2*a** 
3 + 60*sin(c + d*x)**2*a*b**2 + 90*sin(c + d*x)*a**2*b + 60*a**3))/(60*d)